Now, by the principle of homogeneity, i.e., dimensions of LHS and RHS should be equal, hence

$[LHS] = [RHS]$

$$ \Rightarrow $$$[L] = [A] = L$

As $\omega t - kx$ should be dimensionless, $[\omega t] = [kx] = 1$

$$ \Rightarrow $$$[\omega] T = [k] L = 1$

$$ \Rightarrow $$$[\omega] = T^{-1}$ and $[k] = L^{-1}$

Given, $t_1 = 39.6$ s, $t_2 = 39.9$ s and $t_3 = 39.5$ s

Least count of measuring instrument = 0.1 s

(As measurements have only one decimal place)

Precision in the measurement = Least count of the measuring instrument = 0.1 s

Mean value of time for 20 oscillations is given by

$t = \frac{t_1 + t_2 + t_3}{3} = \frac{39.6 + 39.9 + 39.5}{3} = 39.7\text{ s}$

Absolute errors in the measurements

$\Delta t_1 = t - t_1 = 39.7 - 39.6 = 0.1\text{ s}$

$\Delta t_2 = t - t_2 = 39.7 - 39.9 = -0.2\text{ s}$

$\Delta t_3 = t - t_3 = 39.7 - 39.5 = 0.2\text{ s}$

Mean absolute error =

$\frac{|\Delta t_1| + |\Delta t_2| + |\Delta t_3|}{3} = \frac{0.1 + 0.2 + 0.2}{3} = \frac{0.5}{3} \approx 0.2\text{ s} \quad\text{(rounding off up to one decimal place)}$

$\therefore$ Accuracy of measurement = $\pm 0.2\text{ s}$

We know that dimension of energy $=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$

Let $M_1, L_1, T_1$ and $M_2, L_2, T_2$ are units of mass, length and time in given two systems.

$$ \begin{array}{ll} & \therefore M_1=1 \mathrm{~kg}, L_1=1 \mathrm{~m}, T_1=1 \mathrm{~s} \\\\ & M_2=\alpha \mathrm{kg}, L_2=\beta \mathrm{m}, \mathrm{T}_2=\gamma \mathrm{s} \end{array} $$

The magnitude of a physical quantity remains the same, whatever be the system of units of its measurement i.e.,

$$ \begin{aligned} n_1 u_1 & =n_2 u_2 \\ n_2 & =n_1 \frac{u_1}{u_2}=n_1 \frac{\left[\mathrm{M}_1 L_1^2 \mathrm{~T}_1^{-2}\right]}{\left[\mathrm{M}_2 L_2^2 T_2^{-2}\right]}=5\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right] \times\left[\frac{L_1}{\mathrm{~L}_2}\right]^2 \times\left[\frac{\mathrm{T}_1}{T_2}\right]^{-2} \\\\ & =5\left[\frac{1}{\alpha} \mathrm{kg}\right] \times\left[\frac{1}{\beta} \mathrm{m}\right]^2 \times\left[\frac{1}{\gamma} \mathrm{s}\right]^{-2} \\\\ & =5 \times \frac{1}{\alpha} \times \frac{1}{\beta^2} \times \frac{1}{\gamma^{-2}} \\ n_2 & =\frac{5 \gamma^2}{\alpha \beta^2} \end{aligned} $$

Thus, new unit of energy will be $\frac{\gamma^2}{\alpha \beta^2}$.

The volume of a liquid flowing out per second of a pipe is given by $V=\frac{\pi}{8} \frac{p r^4}{\eta l}$

Dimension of $$ V=\frac{\text { Dimension of volume }}{\text { Dimension of time }}=\frac{\left[\mathrm{L}^3\right]}{[T]}=\left[\mathrm{L}^3 \mathrm{~T}^{-1}\right] $$

$(\because V$ is the volume of liquid flowing out per second $)$

Dimension of $p=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$

Dimension of $\eta=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]$

Dimension of $l=[\mathrm{L}]$

Dimension of $r=[\mathrm{L}]$

Dimensions of LHS, $[V]=\frac{\left[L^3\right]}{[T]}=\left[L^3 \mathrm{~T}^{-1}\right]$

Dimensions of RHS, $\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \times\left[\mathrm{L}^4\right]}{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right] \times[\mathrm{L}]}=\left[\mathrm{L}^3 \mathrm{~T}^{-1}\right]$

As dimensions of LHS is equal to the dimensions of RHS. Therefore, equation is correct dimensionally.

Given, physical quantity is $X = a^2 b^3 c^{5/2} d^{-2}$.

Maximum percentage error in $X$ is

$$\frac{\Delta X}{X} \times 100 = \pm \left[ 2 \left(\frac{\Delta a}{a} \times 100 \right) + 3 \left(\frac{\Delta b}{b} \times 100 \right) + \frac{5}{2} \left(\frac{\Delta c}{c} \times 100 \right) + 2 \left(\frac{\Delta d}{d} \times 100 \right) \right]$$

$= \pm \left[2(1) + 3(2) + \frac{5}{2}(3) + 2(4) \right]$ %

$= \pm \left[2 + 6 + 7.5 + 8 \right] = \pm 23.5 \%$

Thus, percentage error in quantity $X$ = ± 23.5%.

Mean absolute error in $X$ = $\pm 0.235 \pm 0.24$ (rounding-off up to two significant digits).

The calculated value of $x$ should be rounded off to two significant digits.

$\therefore X$ = 2.8